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MaRussiya [10]
3 years ago
8

The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it's

shorter. She asks 10 engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should she count on the mean work week to be shorter than 60 hours?Data (length of mean work week): 70; 45; 55; 60; 65; 55; 55; 60; 50; 55.
Mathematics
1 answer:
Sergeu [11.5K]3 years ago
4 0

Answer:

There is evidence to prove that mean is less than 60 hours.

Step-by-step explanation:

Given that the mean work week for engineers in a start-up company is believed to be about 60 hours.

Sample size taken = 10

H_0: \bar x=60\\H_a: \bar x

(left tailed test )

Data (length of mean work week): 70; 45; 55; 60; 65; 55; 55; 60; 50; 55.

\bar x = 57\\s=6.7823\\se = 0.1448

Mean difference = -3

Test statistic t =mean differene/std error

= -20.72

p value <0.00001

Since p < 0.05 at 5% level we reject H0

There is evidence to prove that mean is less than 60 hours.

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In the diagram below AFB is congruent to EFD if EFD = 5x + 6, DFC = 19x - 15, and EFC = 17x + 19, find AFE
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The angle m∠AFE is 128 degrees.

<h3>How to find angles?</h3>

∠AFB ≅ ∠EFD

∠EFD = 5x + 6

m∠DFC = (19x - 15)°

m∠EFC = (17x + 19)°

m∠AFE = ?

m∠AFB + m ∠EFD + m∠AFE = 180

Therefore,

5x + 6 + 5x + 6 + m∠AFE = 180

5x + 5x + 6 + 6 + m∠AFE = 180

10x + 12 + m∠AFE = 180

10x + m∠AFE = 180 - 12

10x + m∠AFE  = 168

m∠AFE = 168 - 10x

m∠EFC =  m ∠EFD + m∠DFC

17x + 19 = 5x + 6 + 19x - 15

17x - 5x - 19x = 6 - 15 - 19

-7x = - 28

x = 28 / 7

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Therefore,

m∠AFE = 168 - 10x

m∠AFE = 168 - 10(4)

m∠AFE = 168 - 40

m∠AFE = 128°

Therefore, the angle m∠AFE = 128°

learn more on angles here: brainly.com/question/13212279

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